Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $y \neq 0$. $p = \dfrac{y^2 - 3y}{-y^2 - 6y + 27} \times \dfrac{y^2 - y - 90}{-2y^2 - 18y} $
Explanation: First factor out any common factors. $p = \dfrac{y(y - 3)}{-(y^2 + 6y - 27)} \times \dfrac{y^2 - y - 90}{-2y(y + 9)} $ Then factor the quadratic expressions. $p = \dfrac {y(y - 3)} {-(y - 3)(y + 9)} \times \dfrac {(y + 9)(y - 10)} {-2y(y + 9)} $ Then multiply the two numerators and multiply the two denominators. $p = \dfrac {y(y - 3) \times (y + 9)(y - 10) } { -(y - 3)(y + 9) \times -2y(y + 9)} $ $p = \dfrac {y(y + 9)(y - 10)(y - 3)} {2y(y - 3)(y + 9)(y + 9)} $ Notice that $(y - 3)$ and $(y + 9)$ appear in both the numerator and denominator so we can cancel them. $p = \dfrac {y(y + 9)(y - 10)\cancel{(y - 3)}} {2y\cancel{(y - 3)}(y + 9)(y + 9)} $ We are dividing by $y - 3$ , so $y - 3 \neq 0$ Therefore, $y \neq 3$ $p = \dfrac {y\cancel{(y + 9)}(y - 10)\cancel{(y - 3)}} {2y\cancel{(y - 3)}(y + 9)\cancel{(y + 9)}} $ We are dividing by $y + 9$ , so $y + 9 \neq 0$ Therefore, $y \neq -9$ $p = \dfrac {y(y - 10)} {2y(y + 9)} $ $ p = \dfrac{y - 10}{2(y + 9)}; y \neq 3; y \neq -9 $